LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Surface existence from rank-three existence on the cylinder

Statement

Assume AC. Fix a field $k$ , let $X$ be a smooth affine finite-type $k$ -scheme of pure dimension two, and put $Y = X \times_{k} A_{k}^{1}$ . Suppose every triple in $\prod_{i = 1}^{3} {C H}^{i} (Y)$ is realized by a rank-three bundle. Then every $(a_{1}, a_{2}) \in {C H}^{1} (X) \times {C H}^{2} (X)$ is realized by a rank-two bundle on $X$ .

Facts & Assumptions

Given: The data and hypotheses in the statement.

Goal: Establish the stated conclusion.

[F1]

The projection and zero section have $p s = id$ and the cylinder is a smooth affine threefold. Cylinder and zero section of a smooth affine surface

[F2]

Chern classes are natural between smooth schemes. Chern classes of a vector bundle on a smooth scheme

[F3]

A rank-three bundle on a noetherian affine surface splits as a trivial line plus a rank-two bundle. Serre splitting above the dimension

[F4]

Both $X$ and $Y$ are smooth equidimensional finite-type schemes over $k$ , and projection and the zero section are $k$ -morphisms. In this scope splitting off a trivial line does not change Chern coefficients. Chern data survive addition of trivial bundles

Proof

1.1F1F2given

Realize the triple $(p^{*} a_{1}, p^{*} a_{2}, 0)$ by a rank-three bundle $G$ on $Y$ , using the assumed existence. Naturality gives $c_{i} (s^{*} G) = s^{*} p^{*} a_{i} = a_{i}$ for $i = 1, 2$ , since pullback is functorial and $p s = id$ .

2.1F3F4step 1.1∎

By [F3] write $s^{*} G ≅ O_{X} \oplus E$ with $E$ of rank two. By [F4], $c_{i} (E) = c_{i} (s^{*} G) = a_{i}$ for $i = 1, 2$ . Thus $E$ realizes the prescribed pair.

Depends on

Used by · 1

Classical smooth critical-rank classifications

Dependency tree · 3 levels

Arrows run from prerequisites to the items that use them. External source nodes link directly to the cited text. The view shows up to three dependency steps; open a prerequisite to continue.